Given that \(E_{p}(x, \delta)\) from Theorem 1.13 can be written as

\[E_{p}(x, \delta)=\frac{1}{p !} \int_{x}^{x+\delta}(x+\delta-t)^{p} f^{(p+1)}(t) d t\]

show that \(E_{p}(x, \delta)=\delta^{p+1} f^{(p+1)}(\xi) /(p+1)\) ! for some \(\xi \in[x, x+\delta]\).

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P Theorem 1.13 (Taylor). Let f be a function that has p+1 bounded and continuous derivatives in the interval (x, x+6). Then f(x+8)= k=0 Sk f(k) (x) + Ep(x, 8), k! where 1 Ep(x, 8) = (x+8) f(p+1)(t)dt: = p! SP+1 f(p+1)() (p+1)! for some = (x, x + d).